. Find the partial fraction decomposition of the following rational functions

Question

x^3-4x^2+2/(x^2+1)(x^2+7)

 

 

3x^4+3x^3+3x^2-2x+3/x(x^2+1)^2

 

 

13x^2-16x+16/2x^3-x^2-8x+4

Answer 1

First one:

 

x³ - 4x² - 2              Ax + B                 Cx + D

_____________  =  ________   +    _________

(x² + 1)(x² + 7)        x² + 1                x² + 7

 

 

             

                         = (Ax + B)(x² + 7) + (Cx + D)(x² + 1)

                            _________________________

                                    (x² + 1)(x² + 7)

 

 

 

                         =   Ax³ + 7Ax + Bx² + 7B + Cx³ + Cx + Dx² + D

                             _______________________________________

                                                 (x² + 1)(x² + 7)

 

 

Equate numerators. and coefficients

 

x³:             1 = A + C

x² :           -4 = B + D

x  :             0 = 7A + C

cons:         -2 = 7B + D

 

Solve the system of equation.

 

We know that from the 3rd equation,    C = -7A

 

1 = A - 7A

1 = -6A

A = -1/6

 

C = -7(-1/6)

C = 7/6

 

Subtract the last equation from the 2nd equation to eliminate the D terms.

 

-2 = -6B

1/3 = B

 

Substitute this value of B into the second equation to solve for D.

 

-4 = (1/3) + D

-13/3 = D

 

Plug in these values into the decomposition.

 

(-1/6)x + (1/3)          (7/6)x - (13/3)

_____________  +   ______________

      x² + 1                      x² + 7

 

 

 

Do the same for the last two.

Answer 2

Oh Sarah, you will definitely need this in Calculus, especially Calc2 and integral Calculus.

 

Here's the first one:

 

As the factors in the denominator are quadratic, the terms in the numerator must be LINEAR.

 

x^3-4x^2+2/(x^2+1)(x^2+7) = (Ax + B)/(x^2 + 1)    +     (Cx+D)/(x^2 + 7)

 

for undetermined coefficients A,B,C and D

 

Multiplying both sides by the common denominator (x^2+1)(x^2+7):

 

x^3-4x^2+2 = (Ax+B)(x^2+7) + (Cx+D)(x^2+1)

 

                      =  Ax^3 + Bx^2 + 7Ax + 7B + Cx^3 + Dx^2 +Cx + D

 

                      =  (A+C)x^3 + (B+D)x^2 + (7A+C)x + (7B+D)

 

 Equating the coefficients:

  1 = A + C  --->  C = 1 - A

 -4 = B + D   --->  D = -4 - B

 0 = 7A + C 

 2 = 7B + D

 

 Substituting the first equation into the 3rd:     0 = 7A + (1 - A) 

                                                                         0 = 6A + 1

                                                                         -1 = 6A

                                                                           A = -1/6 ----> C = 1 - (-1/6) = 6/6 - (-1/6) = 6/6 + 1/6 = 7/6

 

 Substituting the second equation into the 4th:

           7B + -4 - B = 2

 

           6B - 4 = 2

          6B = 6

         B = 1 ---> D = -4 - 1 = -5

 

  So the partial fraction decomposition APPEARS to be:

 

    (-1/6X +1)/(x^+1) + (7/6X-5)/(x^2+7)

 

CHECK:

 

   (-1/6X +1)(x^2+7) + (7/6X-5)(x^2+1)

 

  -1/6X^3 +  X^2 - 7/6X  +  7    +   7/6X^3 - 5X^2 + 7/6X - 5

 

      

  X^3 - 4X^2 +2  which is the original numerator.

 

I will post the solution to the second and third problems in separate comments. Please try them on your own!

Notice in the second problem, the decomposition will be A/x + (Bx+C)/(x^2 +1) +  (Dx+E)/(x^2+1)^2

 

The denominator in the third problem needs factoring